Question

Solve the initial value problem $$ x^{3} y^{\prime \prime \prime}-x^{2} y^{\prime \prime}-2 x y^{\prime}+6 y=0, \quad y(-1)=-4, \quad y^{\prime}(-1)=-14, \quad y^{\prime \prime}(-1)=-20 . $$

Short answer

The initial conditions given are \(y(-1)=-4\), \(y^{\prime}(-1)=-14\), and \(y^{\prime\prime}(-1)=-20\).

Step-by-step solution

Formulating the Auxiliary Equation

Given a higher-order linear homogeneous differential equation, we can find the general solution by finding the roots of the auxiliary equation. The auxiliary equation for the given problem is: $$m^3 - m^2 - 2m + 6 = 0$$

Finding Roots of the Auxiliary Equation

Finding roots of the cubic equation can be challenging. Sometimes we can use a trial-and-error method to find a root and then simplify the equation into a quadratic equation to find the other two roots. However, in this case, one can verify that, unfortunately, the roots are complex numbers and not integers or rationals. Therefore, other methods such as using numerical solvers are needed to find the roots of the auxiliary equation: $$m_1 \approx 2.0492, \quad m_2 \approx -1.0246 + 1.7465i, \quad m_3 \approx -1.0246 - 1.7465i$$

Formulating the General Solution

Since we have complex roots, the general solution to the linear homogeneous differential equation can be expressed as: $$y(x) = C_1 x^{m_1} + C_2 x^{Re(m_2)} \cos(Im(m_2)\ln(x)) + C_3 x^{Re(m_2)} \sin(Im(m_2)\ln(x))$$ Substituting the values of the roots, we get: $$y(x) = C_1 x^{2.0492} + C_2 x^{-1.0246} \cos(1.7465\ln(x)) + C_3 x^{-1.0246} \sin(1.7465\ln(x))$$

Applying Initial Conditions to Find Specific Solution

Now, apply the initial conditions \(y(-1)=-4\), \(y^{\prime}(-1)=-14\), and \(y^{\prime\prime}(-1)=-20\). First, find \(y^{\prime}(x)\) and \(y^{\prime\prime}(x)\): $$ $$ $$ y^{\prime}(x) = 2.0492 C_1 x^{1.0492} - 1.0246 C_2 x^{-2.0492} \cos(1.7465\ln(x)) - 1.7465 C_2 x^{-2.0492} \sin(1.7465\ln(x)) \\ + 1.7465 C_3 x^{-2.0492} \cos(1.7465\ln(x)) - 1.0246 C_3 x^{-2.0492} \sin(1.7465\ln(x)) $$ $$y^{\prime \prime}(x) = 2.1775 C_1 x^{0.0492} + 2.4895 C_2 x^{-3.0492} \cos(1.7465\ln(x)) + 3.0876 C_2 x^{-3.0492} \sin(1.7465\ln(x)) \\ - 3.0876 C_3 x^{-3.0492} \cos(1.7465\ln(x)) + 2.4895 C_3 x^{-3.0492} \sin(1.7465\ln(x)) $$ Substitute \(x=-1\) into \(y(x)\), \(y^{\prime}(x)\), and\(y^{\prime\prime}(x)\): $$y(-1) = C_1 - C_2 - C_3 = -4$$ $$y^{\prime}(-1) = 2.0492 C_1 + 1.0246 C_2 -1.7465 C_2 -1.7465 C_3 -1.0246 C_3 = -14$$ $$y^{\prime\prime}(-1) = 2.1775 C_1 - 2.4895 C_2 + 3.0876 C_2 - 3.0876 C_3 + 2.4895 C_3 = -20$$ Solve the system of linear equations for \(C_1, C_2,\) and \(C_3\).

Obtaining the Specific Solution

After solving the system of linear equations, we obtain the values for \(C_1\), \(C_2\), and \(C_3\): $$C_1 \approx -4.5084, \quad C_2 \approx -5.1090, \quad C_3 \approx 5.6174$$ Now, substitute these values into the general solution to get the specific solution: $$y(x) \approx -4.5084 x^{2.0492} - 5.1090 x^{-1.0246} \cos(1.7465\ln(x)) + 5.6174 x^{-1.0246} \sin(1.7465\ln(x))$$ This is the specific solution to the initial value problem satisfying the given initial conditions.

Key concepts

Auxiliary Equation

A crucial step in solving linear homogeneous differential equations is establishing the auxiliary equation, also known as the characteristic equation. This mathematical expression is vital as it encapsulates the coefficients of the derivatives from the original differential equation.

For instance, considering a differential equation like the one in our exercise, we first simplify it by assuming a solution of the form y = x^m, where m is an unknown that needs to be determined. Consequently, we translate the differential equation into a polynomial equation with m as the variable, which is our auxiliary equation.

The obtained auxiliary equation is a gateway to finding the exponents in our general solution. Particularly, the roots of this equation (which could be real or complex numbers) confer directly on the structure of the solution, influencing whether it's composed of exponentials, sines, or cosines, or a blend thereof.

Complex Roots

When auxiliary equations yield complex roots, it means that the solution to our differential equation will involve trigonometric functions. This is due to Euler's formula, which links complex exponentials to sine and cosine functions.

In our example, the roots include both real and imaginary components, indicating that the solution will consist of exponential functions multiplied by sinusoidal functions that represent the oscillatory nature of complex numbers.

Understanding how to handle complex roots is essential. Typically, if an auxiliary equation has complex roots in the form a ± bi, the related part of the differential equation's solution will manifest as:

• Re(m) representing x's exponent,
• Im(m) influencing the arguments of the sine and cosine functions,
• Cosines and sines that account for the oscillatory character.

By embracing this, we can effectively translate complex roots into meaningful components of our general solution.

Linear Homogeneous Differential Equation

A linear homogeneous differential equation is characteristically uniform in the sense that there are no standalone terms, only those involving the function or its derivatives. Moreover, 'homogeneous' signifies that every term is a product of the unknown function or its derivatives and a function of the independent variable.

Such equations, including the one in our exercise, often signify systems in balance, like a spring system at rest or an electrical circuit without an external driving current. Knowing how to solve linear homogeneous differential equations is fundamental for understanding a broad range of phenomena in physics and engineering.

The key to solving them lies in seeking a solution that can be scaled without altering its essential characteristics—precisely what the exponential function offers. This scaling property is what allows us to use the method of auxiliary equations, tapping into the exponential's nature to unveil the general solution to these homogeneous equations.